add: 只出现一次的数字等

src/array/count-the-repetitions.js

@@ -0,0 +1,149 @@
+/**
+ * 判断从 s2 中删除某些字符是否可以变为 s1
+ * @param {string} s1
+ * @param {string} s2
+ * @return {number} 返回可以变为s1的次数
+ */
+export const includesInStr = (s1, s2) => {
+  let i = 0
+  while (s1.length >= s2.length) {
+    for (let n = 0, len = s2.length; n < len; n++) {
+      const tmp = s1.indexOf(s2[n])
+      if (tmp === -1) {
+        return i
+      } else {
+        s1 = s1.substr(tmp + s2[n].length, s1.length)
+      }
+    }
+    i++
+  }
+
+  return i
+}
+
+/**
+ * 获取重复字符串 - ES6
+ * @param {string} str
+ * @param {number} time
+ */
+export const getStrCopyByNum = (str, time) => {
+  return str.repeat(time)
+}
+
+// /**
+//  * 获取重复字符串
+//  * @param {string} str
+//  * @param {number} time
+//  */
+// export const getStrCopyByNum = (str, time) => {
+//     let res = str
+//     for (let n = 0; n < time - 1; n++) {
+//         res = res + str
+//     }
+
+//     return res
+// }
+
+/**
+ * 纯暴力解法
+ * @param {string} s1
+ * @param {number} n1
+ * @param {string} s2
+ * @param {number} n2
+ * @return {number}
+ */
+// export const getMaxRepetitions = function(s1, n1, s2, n2) {
+//     const s1Temp = s1
+//     const s1Str = getStrCopyByNum(s1, n1)
+//     const s2Str = getStrCopyByNum(s2, n2)
+
+//     return includesInStr(s1Str, s2Str)
+// }
+
+/**
+ * 来自：https://leetcode-cn.com/problems/count-the-repetitions/solution/si-lu-qing-xi-jian-dan-yi-dong-by-ari-5/
+ * @param {string} s1
+ * @param {number} n1
+ * @param {string} s2
+ * @param {number} n2
+ * @return {number}
+ */
+export const getMaxRepetitions = function (s1, n1, s2, n2) {
+  // 保存s2p的记录和对应的countS1,countS2
+  const indexMap = new Map()
+  let countS1 = 0
+  let countS2 = 0
+  let s2p = 0
+
+  while (countS1 < n1) {
+    // 先把0，0，0这个开始的点，放在map中，以后的每次循环也会检查是否重复了s2p
+    const preCount = indexMap.get(s2p)
+    if (preCount === undefined) { // 没有就记录
+      indexMap.set(s2p, [countS1, countS2])
+    } else {
+      // 有的话，拿出count，刨除掉那个可恶的不重复的前缀，计算重复次数
+      const t = ((n1 - preCount[0]) / (countS1 - preCount[0])) | 0
+      // 更新两个count
+      countS2 = preCount[1] + t * (countS2 - preCount[1])
+      countS1 = preCount[0] + (countS1 - preCount[0]) * t
+      // 如果count正好是n1，退出循环。如果还有，还要继续走下去
+      // 为了避免重复读取map和计算，计算完重复就把map清除掉，来一手过河拆桥
+      indexMap.clear()
+      if (countS1 === n1) { break }
+    }
+
+    for (let i = 0; i < s1.length; i++) {
+      if (s1[i] === s2[s2p]) {
+        s2p++
+        if (s2p === s2.length) {
+          s2p = 0
+          countS2++
+        }
+      }
+    }
+    countS1++
+  }
+  // 取整
+  return countS2 / n2 | 0
+}
+
+// /**
+//  * 贪心算法
+//  * @param {string} s1
+//  * @param {number} n1
+//  * @param {string} s2
+//  * @param {number} n2
+//  * @return {number}
+//  */
+// // ['aaa',10]和['aa',2]
+
+// // 如果我们找到了，2个‘aaa’能表示'3个‘aa’,
+
+// // 那么一共10个的话，就可以表示15个‘aa’
+
+// // 那么这个题目的返回值应该就是 15/2 取整，（其中2是‘aa’的个数）
+// export const getMaxRepetitions = function(s1, n1, s2, n2) {
+//     const s1Temp = s1
+//     const s2Str = getStrCopyByNum(s2, n2)
+
+//     let s1Count = 1
+//     let maxTime = 0
+//     let maxLength = 0
+
+//     for (let n = 0; n < n1; n++) {
+//         const time = includesInStr(s1, s2Str)
+//         if (time > 0) {
+//             console.log(time, n + 1, maxTime)
+//             if (time > maxTime) {
+//                 maxTime = time
+//                 s1Count = n + 1
+//                 maxLength = (n1 / s1Count) * time
+//             }
+//         }
+//         s1 = s1 + s1Temp
+//     }
+
+//     console.log(maxLength, s1Count)
+
+//     return maxLength / s1Count
+// }